I can't figure out how to use a "default value" when asking the user for input. I want the user to be able to just press Enter and get the default value. Consider the following piece of code, can you help me?
int number;
cout << "Please give a number [default = 20]: ";
cin >> number;
if(???) {
// The user hasn't given any input, he/she has just
// pressed Enter
number = 20;
}
while(!cin) {
// Error handling goes here
// ...
}
cout << "The number is: " << number << endl;
Use std::getline to read a line of text from std::cin. If the line is empty, use your default value. Otherwise, use a std::istringstream to convert the given string to a number. If this conversion fails, the default value will be used.
Here's a sample program:
#include <iostream>
#include <sstream>
#include <string>
using namespace std;
int main()
{
std::cout << "Please give a number [default = 20]: ";
int number = 20;
std::string input;
std::getline( std::cin, input );
if ( !input.empty() ) {
std::istringstream stream( input );
stream >> number;
}
std::cout << number;
}
This works as an alternative to the accepted answer. I would say std::getline is a bit on the overkill side.
#include <iostream>
int main() {
int number = 0;
if (std::cin.peek() == '\n') { //check if next character is newline
number = 20; //and assign the default
} else if (!(std::cin >> number)) { //be sure to handle invalid input
std::cout << "Invalid input.\n";
//error handling
}
std::cout << "Number: " << number << '\n';
}
Here's a live sample with three different runs and inputs.
if(!cin)
cout << "No number was given.";
else
cout << "Number " << cin << " was given.";
I'd be tempted to read the line as a string using getline() and then you've (arguably) more control over the conversion process:
int number(20);
string numStr;
cout << "Please give a number [default = " << number << "]: ";
getline(cin, numStr);
number = ( numStr.empty() ) ? number : strtol( numStr.c_str(), NULL, 0);
cout << number << endl;
Related
I want to check if the input is valid, but when i do run this code I see that it checks only input for charcters. If i input a float number it will take it and going to use like integer without fractional part.
#inclide <iostream>
using namespace std;
...
int n;
cout << "Your input is: "<<endl;
cin >> n;
while (cin.fail()) {
cout << "Error. Number of elements must be integer. Try again: " << endl;
cin.clear();
cin.ignore(256, '\n');
cin >> n;
}
...
`
So, how to make this code see if the input is float?
You can try to convert the input string to a int using a std::istringstream. If it succeeds then check for eof() (after ignoring blank spaces) to see if the whole input was consumed while converting to int. If the whole input was consumed then it was a valid int.
Something a bit like this:
int input_int()
{
int i;
// get the input
for(std::string line; std::getline(std::cin, line);)
{
// try to convert the input to an int
// if at eof() all of the input was converted - must be an int
if(!line.empty() && (std::istringstream(line) >> i >> std::ws).eof())
break;
// try again
std::cout << "Not an integer please try again: " << std::flush;
}
return i;
}
int main()
{
std::cout << "Enter an integer: " << std::flush;
std::cout << "i: " << input_int() << '\n';
}
Building on Raindrop7's solution, here's the full code to do what you need:
#include <cstdio>
#include <iostream>
#include <cmath>
using namespace std;
int main()
{
double m;
cout << "Your input is: "<<endl;
cin >> m;
while (cin.fail() || (m-floor(m)))
{
cout << "Error. Nubmer of elements has to be integer. Try again: " << endl;
cin.clear();
cin.ignore(256, '\n');
cin >> m;
}
int n = (int)m;
return 0;
}
Here's a sample output:
Your input is:
2.7
Error. Nubmer of elements has to be integer. Try again:
erer
Error. Nubmer of elements has to be integer. Try again:
2
The code below should be able to do what you are hoping to achieve:
#inclide <iostream>
using namespace std;
int n;
cout << "Your input is: "<<endl;
while (!(cin >> n) || cin.get() != '\n') {
cout << "Error. Number of elements must be integer. Try again: " << endl;
cin.clear();
cin.ignore(256, '\n');
}
The program asks the user to re-enter an integer if either of the following happens:
If the program is unable to extract an integer from the std::cin stream. (For example, when a character or string is entered by the user)
If, after an integer is extracted successfully, the next character in std::cin is not the new line '\n' character. (For example, when a number with a decimal point like 1.1 is entered, or when an integer followed by a character like 1a is entered.)
double age = 0;
int empAge;
cout << "Please enter your age in integers: ";
cin >> age;
// Input validation for age. To prevent input with a decimal number eg 5.6, 7.8, 90.9, etc
while(cin.fail() || ((int)age != (double)age)){
cout << "\nThat is not a valid option, please try again. ";
cin >> age;
if(cin.fail() || (int)age != double(age)){
cin.clear();
string not_an_int;
cin >> not_an_int;
}
cout << "That is not a valid option, please try again. ";
cin >> age;
}
// Assigns the data in variable of double type age to the variable of int type empAge
age = empAge;
What I need to get done ?
Ask user to enter a integer digit only.
If not an integer give error "That is not a valid option, please try again" and ask user to enter again.
Attached is also an image showing errors my code gives with the inputs used.
What I need to get done ? Ask user to enter a integer digit only. If
not an integer give error "That is not a valid option, please try
again" and ask user to enter again.
I advise you to read this carefully, as it is quite unclear what exactly you are asking.
I've prepared this small snippet for you in case you are asking something like How do I read user input using stdin while validating it is only an integer value?
#include <iostream>
#include <string>
#include <limits>
bool is_natural(const std::string s)
{
return s.find_first_not_of( "0123456789" ) == std::string::npos;
}
int main()
{
std::string input;
while (!(std::cin >> input) || !is_natural(input))
{
std::cout << "That is not a valid option, please try again.\n";
std::cin.clear();
std::cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
}
int age = std::stoi(input);
std::cout << "Age (natural integer value) entered: " << age << "\n";
}
If you want to read only integers in input, this is a possible solution:
#include <iostream>
#include <string>
using namespace std;
int main() {
int age;
unsigned long status;
string myString;
do {
cout << "Please enter your age in integers: ";
getline(cin, myString);
/*
* Return the position of the first character that does not match.
* Else return "string::npos"
*/
status = myString.find_first_not_of("0123456789");
if (status != string::npos)
cout << "That is not a valid option, please try again." << endl;
} while (status != string::npos);
//Convert string to int.
age = stoi(myString);
cout << "Age as integer: " << age << endl;
}
You can use stoi() only in c++ 11 or following:
int num = stoi( str )
Old method:
int num = atoi( str.c_str() )
using namespace std;
// The method check if the string only contains digits
bool is_digits(const std::string &str)
{
return std::all_of(str.begin(), str.end(), ::isdigit);
}
int main()
{
int age;
bool status = false;
string myString;
while (!status) {
cout << "Please enter your age in integers: ";
cin >> myString;
status = is_digits(myString);
if (!status)
cout << "That is not a valid option, please try again." << endl;
}
//Convert string to int.
age = stoi(myString);
cout << "Age as integer: " << age << endl;
return 0;
}
I was typing this and it asks the user to input two integers which will then become variables. From there it will carry out simple operations.
How do I get the computer to check if what is entered is an integer or not? And if not, ask the user to type an integer in. For example: if someone inputs "a" instead of 2, then it will tell them to reenter a number.
Thanks
#include <iostream>
using namespace std;
int main ()
{
int firstvariable;
int secondvariable;
float float1;
float float2;
cout << "Please enter two integers and then press Enter:" << endl;
cin >> firstvariable;
cin >> secondvariable;
cout << "Time for some simple mathematical operations:\n" << endl;
cout << "The sum:\n " << firstvariable << "+" << secondvariable
<<"="<< firstvariable + secondvariable << "\n " << endl;
}
You can check like this:
int x;
cin >> x;
if (cin.fail()) {
//Not an int.
}
Furthermore, you can continue to get input until you get an int via:
#include <iostream>
int main() {
int x;
std::cin >> x;
while(std::cin.fail()) {
std::cout << "Error" << std::endl;
std::cin.clear();
std::cin.ignore(256,'\n');
std::cin >> x;
}
std::cout << x << std::endl;
return 0;
}
EDIT: To address the comment below regarding input like 10abc, one could modify the loop to accept a string as an input. Then check the string for any character not a number and handle that situation accordingly. One needs not clear/ignore the input stream in that situation. Verifying the string is just numbers, convert the string back to an integer. I mean, this was just off the cuff. There might be a better way. This won't work if you're accepting floats/doubles (would have to add '.' in the search string).
#include <iostream>
#include <string>
int main() {
std::string theInput;
int inputAsInt;
std::getline(std::cin, theInput);
while(std::cin.fail() || std::cin.eof() || theInput.find_first_not_of("0123456789") != std::string::npos) {
std::cout << "Error" << std::endl;
if( theInput.find_first_not_of("0123456789") == std::string::npos) {
std::cin.clear();
std::cin.ignore(256,'\n');
}
std::getline(std::cin, theInput);
}
std::string::size_type st;
inputAsInt = std::stoi(theInput,&st);
std::cout << inputAsInt << std::endl;
return 0;
}
Heh, this is an old question that could use a better answer.
User input should be obtained as a string and then attempt-converted to the data type you desire. Conveniently, this also allows you to answer questions like “what type of data is my input?”
Here is a function I use a lot. Other options exist, such as in Boost, but the basic premise is the same: attempt to perform the string→type conversion and observe the success or failure:
template <typename T>
auto string_to( const std::string & s )
{
T value;
std::istringstream ss( s );
return ((ss >> value) and (ss >> std::ws).eof()) // attempt the conversion
? value // success
: std::optional<T> { }; // failure
}
Using the optional type is just one way. You could also throw an exception or return a default value on failure. Whatever works for your situation.
Here is an example of using it:
int n;
std::cout << "n? ";
{
std::string s;
getline( std::cin, s );
auto x = string_to <int> ( s );
if (!x) return complain();
n = *x;
}
std::cout << "Multiply that by seven to get " << (7 * n) << ".\n";
limitations and type identification
In order for this to work, of course, there must exist a method to unambiguously extract your data type from a stream. This is the natural order of things in C++ — that is, business as usual. So no surprises here.
The next caveat is that some types subsume others. For example, if you are trying to distinguish between int and double, check for int first, since anything that converts to an int is also a double.
There is a function in c called isdigit(). That will suit you just fine. Example:
int var1 = 'h';
int var2 = '2';
if( isdigit(var1) )
{
printf("var1 = |%c| is a digit\n", var1 );
}
else
{
printf("var1 = |%c| is not a digit\n", var1 );
}
if( isdigit(var2) )
{
printf("var2 = |%c| is a digit\n", var2 );
}
else
{
printf("var2 = |%c| is not a digit\n", var2 );
}
From here
If istream fails to insert, it will set the fail bit.
int i = 0;
std::cin >> i; // type a and press enter
if (std::cin.fail())
{
std::cout << "I failed, try again ..." << std::endl
std::cin.clear(); // reset the failed state
}
You can set this up in a do-while loop to get the correct type (int in this case) propertly inserted.
For more information: http://augustcouncil.com/~tgibson/tutorial/iotips.html#directly
You can use the variables name itself to check if a value is an integer.
for example:
#include <iostream>
using namespace std;
int main (){
int firstvariable;
int secondvariable;
float float1;
float float2;
cout << "Please enter two integers and then press Enter:" << endl;
cin >> firstvariable;
cin >> secondvariable;
if(firstvariable && secondvariable){
cout << "Time for some simple mathematical operations:\n" << endl;
cout << "The sum:\n " << firstvariable << "+" << secondvariable
<<"="<< firstvariable + secondvariable << "\n " << endl;
}else{
cout << "\n[ERROR\tINVALID INPUT]\n";
return 1;
}
return 0;
}
I prefer to use <limits> to check for an int until it is passed.
#include <iostream>
#include <limits> //std::numeric_limits
using std::cout, std::endl, std::cin;
int main() {
int num;
while(!(cin >> num)){ //check the Input format for integer the right way
cin.clear();
cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
cout << "Invalid input. Reenter the number: ";
};
cout << "output= " << num << endl;
return 0;
}
Under C++11 and later, I have the found the std::stoi function very useful for this task. stoi throws an invalid_argument exception if conversion cannot be performed. This can be caught and handled as shown in the demo function 'getIntegerValue' below.
The stoi function has a second parameter 'idx' that indicates the position of the first character in the string after the number. We can use the value in idx to check against the string length and ascertain if there are any characters in the input other than the number. This helps eliminate input like 10abc or a decimal value.
The only case where this approach fails is when there is trailing white space after the number in the input, that is, the user enters a lot of spaces after inputting the number. To handle such a case, you could rtrim the input string as described in this post.
#include <iostream>
#include <string>
bool getIntegerValue(int &value);
int main(){
int value{};
bool valid{};
while(!valid){
std::cout << "Enter integer value: ";
valid = getIntegerValue(value);
if (!valid)
std::cout << "Invalid integer value! Please try again.\n" << std::endl;
}
std::cout << "You entered: " << value << std::endl;
return 0;
}
// Returns true if integer is read from standard input
bool getIntegerValue(int &value){
bool isInputValid{};
int valueFromString{};
size_t index{};
std::string userInput;
std::getline(std::cin, userInput);
try {
//stoi throws an invalid_argument exception if userInput cannot be
//converted to an integer.
valueFromString = std::stoi(userInput, &index);
//index being different than length of string implies additional
//characters in input that could not be converted to an integer.
//This is to handle inputs like 10str or decimal values like 10.12
if(index == userInput.length()) isInputValid = true;
}
catch (const std::invalid_argument &arg) {
; //you could show an invalid argument message here.
}
if (isInputValid) value = valueFromString;
return isInputValid;
}
You could use :
int a = 12;
if (a>0 || a<0){
cout << "Your text"<<endl;
}
I'm pretty sure it works.
I was typing this and it asks the user to input two integers which will then become variables. From there it will carry out simple operations.
How do I get the computer to check if what is entered is an integer or not? And if not, ask the user to type an integer in. For example: if someone inputs "a" instead of 2, then it will tell them to reenter a number.
Thanks
#include <iostream>
using namespace std;
int main ()
{
int firstvariable;
int secondvariable;
float float1;
float float2;
cout << "Please enter two integers and then press Enter:" << endl;
cin >> firstvariable;
cin >> secondvariable;
cout << "Time for some simple mathematical operations:\n" << endl;
cout << "The sum:\n " << firstvariable << "+" << secondvariable
<<"="<< firstvariable + secondvariable << "\n " << endl;
}
You can check like this:
int x;
cin >> x;
if (cin.fail()) {
//Not an int.
}
Furthermore, you can continue to get input until you get an int via:
#include <iostream>
int main() {
int x;
std::cin >> x;
while(std::cin.fail()) {
std::cout << "Error" << std::endl;
std::cin.clear();
std::cin.ignore(256,'\n');
std::cin >> x;
}
std::cout << x << std::endl;
return 0;
}
EDIT: To address the comment below regarding input like 10abc, one could modify the loop to accept a string as an input. Then check the string for any character not a number and handle that situation accordingly. One needs not clear/ignore the input stream in that situation. Verifying the string is just numbers, convert the string back to an integer. I mean, this was just off the cuff. There might be a better way. This won't work if you're accepting floats/doubles (would have to add '.' in the search string).
#include <iostream>
#include <string>
int main() {
std::string theInput;
int inputAsInt;
std::getline(std::cin, theInput);
while(std::cin.fail() || std::cin.eof() || theInput.find_first_not_of("0123456789") != std::string::npos) {
std::cout << "Error" << std::endl;
if( theInput.find_first_not_of("0123456789") == std::string::npos) {
std::cin.clear();
std::cin.ignore(256,'\n');
}
std::getline(std::cin, theInput);
}
std::string::size_type st;
inputAsInt = std::stoi(theInput,&st);
std::cout << inputAsInt << std::endl;
return 0;
}
Heh, this is an old question that could use a better answer.
User input should be obtained as a string and then attempt-converted to the data type you desire. Conveniently, this also allows you to answer questions like “what type of data is my input?”
Here is a function I use a lot. Other options exist, such as in Boost, but the basic premise is the same: attempt to perform the string→type conversion and observe the success or failure:
template <typename T>
auto string_to( const std::string & s )
{
T value;
std::istringstream ss( s );
return ((ss >> value) and (ss >> std::ws).eof()) // attempt the conversion
? value // success
: std::optional<T> { }; // failure
}
Using the optional type is just one way. You could also throw an exception or return a default value on failure. Whatever works for your situation.
Here is an example of using it:
int n;
std::cout << "n? ";
{
std::string s;
getline( std::cin, s );
auto x = string_to <int> ( s );
if (!x) return complain();
n = *x;
}
std::cout << "Multiply that by seven to get " << (7 * n) << ".\n";
limitations and type identification
In order for this to work, of course, there must exist a method to unambiguously extract your data type from a stream. This is the natural order of things in C++ — that is, business as usual. So no surprises here.
The next caveat is that some types subsume others. For example, if you are trying to distinguish between int and double, check for int first, since anything that converts to an int is also a double.
There is a function in c called isdigit(). That will suit you just fine. Example:
int var1 = 'h';
int var2 = '2';
if( isdigit(var1) )
{
printf("var1 = |%c| is a digit\n", var1 );
}
else
{
printf("var1 = |%c| is not a digit\n", var1 );
}
if( isdigit(var2) )
{
printf("var2 = |%c| is a digit\n", var2 );
}
else
{
printf("var2 = |%c| is not a digit\n", var2 );
}
From here
If istream fails to insert, it will set the fail bit.
int i = 0;
std::cin >> i; // type a and press enter
if (std::cin.fail())
{
std::cout << "I failed, try again ..." << std::endl
std::cin.clear(); // reset the failed state
}
You can set this up in a do-while loop to get the correct type (int in this case) propertly inserted.
For more information: http://augustcouncil.com/~tgibson/tutorial/iotips.html#directly
You can use the variables name itself to check if a value is an integer.
for example:
#include <iostream>
using namespace std;
int main (){
int firstvariable;
int secondvariable;
float float1;
float float2;
cout << "Please enter two integers and then press Enter:" << endl;
cin >> firstvariable;
cin >> secondvariable;
if(firstvariable && secondvariable){
cout << "Time for some simple mathematical operations:\n" << endl;
cout << "The sum:\n " << firstvariable << "+" << secondvariable
<<"="<< firstvariable + secondvariable << "\n " << endl;
}else{
cout << "\n[ERROR\tINVALID INPUT]\n";
return 1;
}
return 0;
}
I prefer to use <limits> to check for an int until it is passed.
#include <iostream>
#include <limits> //std::numeric_limits
using std::cout, std::endl, std::cin;
int main() {
int num;
while(!(cin >> num)){ //check the Input format for integer the right way
cin.clear();
cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
cout << "Invalid input. Reenter the number: ";
};
cout << "output= " << num << endl;
return 0;
}
Under C++11 and later, I have the found the std::stoi function very useful for this task. stoi throws an invalid_argument exception if conversion cannot be performed. This can be caught and handled as shown in the demo function 'getIntegerValue' below.
The stoi function has a second parameter 'idx' that indicates the position of the first character in the string after the number. We can use the value in idx to check against the string length and ascertain if there are any characters in the input other than the number. This helps eliminate input like 10abc or a decimal value.
The only case where this approach fails is when there is trailing white space after the number in the input, that is, the user enters a lot of spaces after inputting the number. To handle such a case, you could rtrim the input string as described in this post.
#include <iostream>
#include <string>
bool getIntegerValue(int &value);
int main(){
int value{};
bool valid{};
while(!valid){
std::cout << "Enter integer value: ";
valid = getIntegerValue(value);
if (!valid)
std::cout << "Invalid integer value! Please try again.\n" << std::endl;
}
std::cout << "You entered: " << value << std::endl;
return 0;
}
// Returns true if integer is read from standard input
bool getIntegerValue(int &value){
bool isInputValid{};
int valueFromString{};
size_t index{};
std::string userInput;
std::getline(std::cin, userInput);
try {
//stoi throws an invalid_argument exception if userInput cannot be
//converted to an integer.
valueFromString = std::stoi(userInput, &index);
//index being different than length of string implies additional
//characters in input that could not be converted to an integer.
//This is to handle inputs like 10str or decimal values like 10.12
if(index == userInput.length()) isInputValid = true;
}
catch (const std::invalid_argument &arg) {
; //you could show an invalid argument message here.
}
if (isInputValid) value = valueFromString;
return isInputValid;
}
You could use :
int a = 12;
if (a>0 || a<0){
cout << "Your text"<<endl;
}
I'm pretty sure it works.
When running the following code and enter a number, it works fine.
But when entering a letter, the program enters an infinite loop, displaying "Enter a number (0 to exit): cin failed."
My intent was to handle the cin fail case and prompt the user again.
int number;
do{
cout << "Enter a number (0 to exit): ";
cin >> number;
if(cin.fail()){
cout << "cin failed." << endl;
cin.clear();
}else{
cout << "cin succeeded, " << number << " entered." << endl;
}
}while(number != 0);
You need to clear the line from cin, using cin.ignore, in addition to clearing the stream state (which is what cin.clear does).
I have several utility functions to make this easier (you'll be interested in clearline in particular, which clears the stream state and the current line) and almost an exact example of what you want.
Your code, more or less, using my clearline:
#include "clinput.hpp" // move my file to a location it can be used from
int main() {
using namespace std;
while (true) {
cout << "Enter a number (0 to exit): ";
int number;
if (cin >> number) {
cout << "Read " << number << '\n';
if (number == 0) {
break;
}
}
else {
if (cin.eof()) { // tested only *after* failed state
cerr << "Input failed due to EOF, exiting.\n";
return 1;
}
cerr << "Input failed, try again.\n";
clearline(cin); // "cin >> clearline" is identical
}
}
return 0;
}
There is still a potential issue here (fixed in my clinput_loop.cpp with blankline), with leaving input in the buffer that will screw up later IO (see "42 abc" in the sample session). Extracting the above code into a separate and self-contained function is left as an exercise for the reader, but here's a skeleton:
template<class Type, class Ch, class ChTr>
Type read(std::basic_istream<Ch,ChTr>& stream, Ch const* prompt) {
Type value;
// *try input here*
if (could_not_get_input or more_of_line_left) {
throw std::runtime_error("...");
}
return value;
}
template<class Type, class Ch, class ChTr>
void read_into(
Type& value,
std::basic_istream<Ch,ChTr>& stream,
Ch const* prompt
) {
value = read<Type>(stream, prompt);
}
Example use:
int n;
try {
read_into(n, std::cin, "Enter a number: ");
}
catch (std::runtime_error& e) {
//...
raise;
}
cout << "Read " << n << '\n';
clearline function extracted for posterity, in case above links ever break (and slightly changed to make self-contained):
#include <istream>
#include <limits>
template<class C, class T>
std::basic_istream<C,T>& clearline(std::basic_istream<C,T>& s) {
s.clear();
s.ignore(std::numeric_limits<std::streamsize>::max(), s.widen('\n'))
return s;
}
The template stuff is a bit confusing if you're not used to it, but it's not hard:
std::istream is a typedef of std::basic_istream<char, std::char_traits<char> >
std::wistream is a typedef of std::basic_istream<wchar_t, std::char_traits<wchar_t> >
widen allows '\n' to become L'\n' as appropriate
this code works for both of the common char and wchar_t cases, but also any compatible instantiation of basic_istream
it's written to be called as clearline(stream) or stream >> clearline, compare to other manipulators like std::endl, std::ws, or std::boolalpha
This is probably what you intended to do:
#include <iostream>
using namespace std;
int main ()
{
int i;
do {
if (cin.fail())
{
cin.ignore(255);
cin.clear();
}
cout << "Please enter an integer value: ";
cin >> i;
} while ( cin.fail() );
cout << "The value you entered is " << i;
return 0;
}
This is simple example of cin.fail()
It will process input until a valid integer value is provided
#include <iostream>
using namespace std;
int main()
{
int j;
int i;
i = 0;
while (1) {
i++;
cin >> j;
if (cin.fail()) return 0;
cout << "Integer " << i << ": " << j << endl;
}
}
Input:
42 51 85 hello 85
Output:
Integer 1: 42
Integer 2: 51
Integer 3: 85